CONTEXT
I am self studying Hilbert Space theory, and since I am still moving the first steps into the field, I am not quite sure of the conclusions I draw while working on the exercises.
The question is related to Exercise 4, Chapter 3 of Debnath and Mikusinski's "Introduction to Hilbert Spaces".
STATEMENT OF THE PROBLEM
Let $\mathcal C^1([a,b])$ be the set of all continuously differentiable complex-valued functions on $[a,b]$ and $\mathcal F = \{f \in \mathcal C^1([a,b]): \ f(a) = 0\}.$
I already showed that $\mathcal F$ is an inner product space with inner product defined by $$ \langle f,g \rangle = \int_a^b f'(x) \overline{g'(x)} dx.\tag{1}\label{1}$$
Now I need to show that this is not a Hilbert space.
In order to do so, I considered, in the interval $[0,1]$, the sequence of functions $$f_n(x) = \int_0^x g_n(t) dt,$$ where, in turn, for $n=1,2,\dots$,
$$g_n(x) = \begin{cases} 0 & \left(x \leq \frac12\right)\\ 2n\left(x-\frac12\right) & \left(\frac12 < x \leq \frac12 + \frac1{2n}\right)\\ 1 & \left(x > \frac12 + \frac1{2n}\right).\end{cases}$$
I want to show that this is a Cauchy sequence that does not converge to any function in $\mathcal F$.
MY QUESTIONS
- In order to show that $(f_n)$ is a Cauchy sequence, I explicitely determined, for $m>n>0$,$$||f_m(x)-f_n(x)||^2 = \int_0^1 |g_m(x) - g_n(x)|^2 dx = \frac{(m-n)^2}{6m^2n}< \frac1{6n}.$$ But I have a feeling this is not really necessary. Can I use any convergence theorem, instead, even if I am not working in $L^1([0,1])$?
- Clearly $(f_n(x))$ converges pointwise to $$f(x) = \begin{cases}0 & \left(x \leq \frac12\right)\\ x-\frac12 & \left(x > \frac12\right),\end{cases}$$which is not in $\mathcal C^1([0,1])$ and thus not in $\mathcal F$. Is this sufficient to conclude that the sequence $(f_n)$ does not converge in $\mathcal F$ with the norm induced by \eqref{1}?
No convergence theorem is directly applicable for showing that $(f_n)$ is Cauchy. Estimating $\|f_n-f_m\|$ is the right way to prove this.
Suppose $f_n \to g$ in $C^{1}$. Then $f_n' \to g'$ in $L^{2}$. Since $f_n(x)=\int_0^{x} f_n'(t)dt$ and $g(x)=\int_0^{x} g'(t)dt$ we see that $f_n \to g$ point-wise, in fact uniformly. [Use Cauchy-Schwarz inequality]. Hence we must have $g(x)=f(x)$ for all $x$. But $f$ is not differentiable at $\frac 1 2$ and $g$ is differentiable at $\frac 1 2$ leadint to a contradiction.